Surface area of revolution of a parametric curve, horizontal axis

Formula for the surface area of revolution

The surface area of the solid created by revolving a parametric curve around the $x$-axis is given by

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$S_x=\int^b_a 2\pi{y}\sqrt{\left[f'(t)\right]^2+\left[g'(t)\right]^2}\ dt$

where the curve is defined over the interval $[a,b]$,

where $f'(t)$ is the derivative of the curve $f(t)$

where $g'(t)$ is the derivative of the curve $g(t)$

How to find the surface area of revolution of a parametric curve

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Finding surface area of revolution of a parametric curve when we rotate around the x-axis

Example

Find the surface area of revolution of the solid created when the parametric curve is rotated around the $x$-axis over the interval $0\le t\le\frac{\pi}{2}$.

$x=\cos^3{t}$

$y=\sin^3{t}$

We’ll call the parametric equations

$f(t)=\cos^3{t}$

$g(t)=\sin^3{t}$

The limits of integration are defined in the problem, but we need to find both derivatives before we can plug into the formula.

$f'(t)=-3\cos^2{t}\sin{t}$

$g'(t)=3\sin^2{t}\cos{t}$

Now we’ll plug into the formula for the surface area of revolution.

$S_x=\int^{\frac{\pi}{2}}_0 2\pi{\left(\sin^3{t}\right)}\sqrt{\left(-3\cos^2{t}\sin{t}\right)^2+\left(3\sin^2{t}\cos{t}\right)^2}\ dt$

$S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\cos^4{t}\sin^2{t}+9\sin^4{t}\cos^2{t}}\ dt$

$S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\sin^2{t}\cos^2{t}\left(\cos^2{t}+\sin^2{t}\right)}\ dt$

We need to find both derivatives before we can plug into the formula.

Since $\sin^2{t}+\cos^2{t}=1$, we get

$S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\sin^2{t}\cos^2{t}\left(1\right)}\ dt$

$S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\sqrt{9\sin^2{t}\cos^2{t}}\ dt$

$S_x=\int^{\frac{\pi}{2}}_0 2\pi\sin^3{t}\left(3\sin{t}\cos{t}\right)\ dt$

$S_x=6\pi\int^{\frac{\pi}{2}}_0\sin^4{t}\cos{t}\ dt$

We’ll use u-substitution, letting

$u=\sin{t}$

$du=\cos{t}\ dt$

We’ll make the substitution.

$S_x=6\pi\int^{x=\frac{\pi}{2}}_{x=0}u^4\ du$

$S_x=\frac{6\pi}{5}u^5\Big|^{x=\frac{\pi}{2}}_{x=0}$

Back-substituting for $u$, we get

$S_x=\frac{6\pi}{5}\sin^5{t}\Big|^{\frac{\pi}{2}}_0$

$S_x=\left(\frac{6\pi}{5}\sin^5{\frac{\pi}{2}}\right)-\left(\frac{6\pi}{5}\sin^5{0}\right)$

$S_x=\frac{6\pi}{5}(1)^5-\frac{6\pi}{5}(0)^5$

$S_x=\frac{6\pi}{5}$

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